Mathematics

A geometric progression builds by multiplying by the same factor each time, so its sum grows in a predictable way. You'll use the geometric series sum formula $S_n = a\frac{r^n - 1}{r - 1}$ to turn "sum equals 484" into an equation for $n$. The key insight is that once you substitute your known values (first term and common ratio), you'll get a simple exponential equation—solve it by recognizing powers of 3.

A geometric progression builds by multiplying by the same factor each time, so its sum grows in a predictable way. You'll use the geometric series sum formula $S_n = a\frac{r^n - 1}{r - 1}$ to turn "sum equals 484" into an equation for $n$. The key insight is that once you substitute your known values (first term and common ratio), you'll get a simple exponential equation—solve it by recognizing powers of 3.